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Absolute Primes

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Absolute Primes Absolute Primes A. Slinko For a long time primes have attracted the attention of mathematicians, especially those primes that possess some sort of symmetry. The mysterious and inconceivable repunits An = 111 . 1(n), whose decimal representation contains only units, form an important class of 1 them . For a repunit An to be prime, the number n of digits in its decimal representation must be also prime. But this condition is far from being sufficient: for instance, A = 111 = 3 37 and A = 11111 = 41 271. Some 3 · 5 · repunits are nonetheless prime: A2, A19, A23, A317 and possibly A1031, give the examples. The question of primeness of repunits was discussed by M. Gardner [1] and later in [2-4]. It is completely unclear whether the number of prime repunits is finite or infinite. The prime repunits are examples of integers which are prime and remain prime after an arbitrary permutation of their decimal digits. Integers with this property are called either permutable primes according to H.-E. Richert [5], who introduced them some 40 years ago, or absolute primes according to T.N. Bhagava and P.H. Doyle [6], and A.W.Johnson [7]. The intent of this note is to give a short proof, which does not require significant num- ber crunching, of all known facts referring to absolute primes different from repunits. Analyzing the table of primes which are less than 103, we find 21 absolute primes different form the repunit 11: 2, 3, 5, 7, 13, 17, 31, 37, 71, 73, 79, 97, 113, 131, 199, 311, 337, 373, 733, 919, 991. 1By a subscript in brackets we will indicate the number of digits in the decimal repre- sentation of the integer. 1 The first observation is easy to get: Lemma 1: A multidigit absolute prime contains in its decimal represen- tations only the four digits 1, 3, 7, 9. Proof: If any of the digits 0, 2, 4, 5, 6, 8 appear in the representation of an integer, then by shifting this digit to the units place we get a multiple of 2 or a multiple of 5. Now we can confine the area of the search, and this helps us to proceed with the following deliberations. Lemma 2: An absolute prime does not contain in its decimal represen- tation all of the digits 1, 3, 7, 9 simultaneously. Proof: Let N be a number with all of the digits 1, 3, 7, 9 in its decimal representation. Let us shift these four digits to the rightmost four places, to obtain an integer 5 N0 = c1 . cn 47931 = L 10 + 7931, − · n 1 n 2 where the notation a1 . an is used to denote the number a110 − +a210 − + . + an 110 + an, which decimal representation consists of digits a1, . , an. − The integers K0 = 7931, K1 = 1793, K2 = 9137, K3 = 7913, K4 = 7193, K5 = 1937, K6 = 7139 have different remainders on dividing by 7 since 5 Ki i (mod 7). The seven integers Ni = L 10 + Ki for i = 0, 1, 2, 3, 4, 5, 6 also≡have different remainders on dividing b·y 7. Therefore one of them is a multiple of 7. Since these integers can be obtained from N by a permutation of digits, N is not an absolute prime. Lemma 3: No absolute prime contains in its decimal representation three digits a and two digits b simultaneously, provided a = b. 6 Proof: Suppose that an integer N contains digits a, a, a, b, b in its decimal representation. By a permutation of digits of N, we can obtain integers i j Ni,j = c1 . cn 5aaaaa + (b a)(10 + 10 ), − − where 4 i > j 0. Since the integers 104 + 101, 103 + 102, 103 + 101, 102 + 100, 101 +≥ 100, 10≥4 + 100, 104 + 102 yield different remainders on dividing by 7, which are respectively 0, 1, 2, 3, 4, 5, 6, so do the integers (b a)(10i + 10j), − when 4 i > j 0. Therefore among the integers Ni,j there exists one that is a multiple≥ of 7.≥ 2 Using these two lemmas we are able by direct calculation (preferably on a computer) to prove that no n-digit absolute primes exist with n = 4, 5, 6. For example, if n = 6, we have to check all the numbers aaaaab, aaaabc, aabbcc, where a, b, c 1, 3, 7, 9 . ∈ { } Lemma 4: If N = c1 . cn 6aaaaab is an absolute prime, then K = − c1 . cn 6 is divisible by 7. − Proof: By permutations of the last 6 digits of N we can obtain the integers N = K 106 + a A + (b a)10i i · · 6 − for 0 i 5. Since b a is even and the powers 10i, 0 i 5, have different nonzero≤ remainders≤ on− dividing by 7: ≤ ≤ 100 1, 101 3, 102 2, 103 6, 104 4, 105 5, ≡ ≡ ≡ ≡ ≡ ≡ i 6 the integers (b a)10 have the same property. If the integer K 10 + a A6 had a nonzero remainder− on dividing by 7, we would find an integer· (b a)10· i, − which has the opposite remainder, and get Ni which is divisible by 7. Since 6 this is impossible, the number K 10 + a A6 is a multiple of 7. Moreover, as 0 1 2 3 4 · 5 · A6 = 10 +10 +10 +10 +10 +10 1+3+2+6+4+5 = 21 0 (mod 7), we conclude that K 106 and hence K≡, is divisible by 7. ≡ · Theorem 1: Every multidigit absolute prime integer N is either a re- punit, or can be obtained by a permutations of digits from the integer B (a, b) = aaa . ab = a A + (b a), n (n) · n − where a and b are different digits from the set 1, 3, 7, 9 . Proof: Let n be the number of digits of N.{We can}suppose that n > 6. By the first three lemmas N is written by the digits 1, 3, 7, 9 only but it does not contain in its decimal representation all of the digits 1, 3, 7, 9, and it can contain three such digits only if N is a permutation of digits of the number aaa . abc(n). Let us show that this is impossible. Since N is an absolute prime, the integers N1 = a . acaaaaab(n), N2 = a . abaaaaac(n), 3 are also absolute primes, and by Lemma 4 the integers a . ac(n 6) and − a . ab(n 6) are both divisible by 7. Thus their difference, whose absolute value is −b c , is also divisible by 7, which is impossible. Therefore| − |either N is a repunit or else it is written by two digits only. In the latter case we need Lemma 3 again, to secure that one digit appears only once. The prime number 7 played a significant role in the preceding consider- ations. But other useful primes also exist and we are going to find some of them. Note that the property of 7 most useful for us was the fact that the powers 10i, 0 < i < 6, had different nonzero remainders on dividing by 7. In general, by Fermat’s Little Theorem for every prime p > 5, we have p 1 10 − 1 (mod p). Let≡h(p) be the least possible positive integer such that 10h(p) 1 (mod p). It is obvious that h(p) is a divisor of p 1 and that 10q 1 ≡(mod p) im- plies that q is divisible by h(p). It is also− easy to see that≡the powers 10j, 0 < j < p 1, have different nonzero remainders on dividing by p as soon as h(p) = p− 1. When this is the case, 10 is said to be a primitive root modulo p. − Note that 10 is a primitive root modulo primes 17, 19, 23, 29 (the reader again may write a computer program to check that), but 10 is not a primitive root modulo 13 since 106 1 (mod 13). ≡ Lemma 5: Let An be a repunit and p > 3 is a prime. Then An 0 (mod p) if and only if n 0 (mod h(p)). ≡ n ≡ Proof: As 10 = 9 An + 1, we have An 0 (mod p) if and only if 10n 1 (mod p) and this· is equivalent to n 0 ≡(mod h(p)). ≡ ≡ This simple assertion gives information about divisors of the repunits: in particular, if n is prime and An = p1p2 . ps is the factorization of An into prime factors, then h(p1) = h(p2) = . = h(ps) = n. For instance, A = 239 4649 and h(239) = h(4649) = 7. 7 · Lemma 6: Let Bn(a, b) be an absolute prime, p be a prime such that n > p 1. Suppose that 10 is a primitive root modulo p, and a and p are relatively− prime. Then n is a multiple of p 1. Proof: Let us consider the integers − p 1 i Bi = a 10 − An p+1 + a Ap 1 + (b a) 10 , 0 i p 2, · · − · − − · ≤ ≤ − 4 obtained from Bn(a, b) by a permutation of the last p 1 digits. The powers 10i, 0 i p 2, yield all nonzero remainders on−dividing by p, so do ≤ ≤ − i the integers (b a) 10 , 0 i p 2, and hence all the integers Bi can − · ≤ ≤ − p 1 be simultaneously prime only in the case when the integer L = a 10 − p 1 · · An p+1 + a Ap 1 is divisible by p. But then, since GCD(a 10 − , p) = 1 and − · − · Ap 1 0 (mod p), it follows that An p+1 is divisible by p and by Lemma 5 n is− divisible≡ by p 1.
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